Post on 20-May-2018
Name___________________________ IGCSE-Hooke’s Law Worksheet Answers
1.1.
1.2.
1 | P a g e
F = k . x
Therefore k = F/x…..
k1 = k2 therefore
F1/x1 =F2/x2
100/10 = m /20
10 m = 2000…. M = 200g
Or
If 100g causes an extention of 10cm than 200g should cause an extension of 20cm
X = 20cm -10cm = 10cm
Spring 1 X = 30cm -10cm = 200cm
Spring 2
ans
ans
1.3.
1.4.
2 | P a g e
15.2 cm + 2.1 = 17.3cm
Mass is the same
Weight is the same
Volume is different
Density is different
ans
1.5.
1.6.
3 | P a g e
From graph, 400N on planet P has a mass of 40 kg.
Mass does not change.
From the graph 40kg has a mas of 200N in planet Q
ans
2.1.
4 | P a g e
X = 16.4 cm – 15 cm
X= 1.4cm
From graph an extension of 1.4 =3.5 N
ans
F = 9N – 7N (opp. Direction- subtract)
= 2N left
2N left
F = m x a
a = F/m = 2/ 0.5kg = 4m.s-2
5 | P a g e
X = 1
X = 1
X = 1
X = 1
X = 1
X = 1
X = 1
X = 0.8
X = 0.8
X = 0.8
X = 1
X= 1.2
X = 1.4
X=1.6
X = 0.8
X = 0.8
X = 0.8
X = 1
X= 1.2
X = 1.4
X=1.6
X = 1.1
X = 1.1
X = 1.1
X = 1.1
X= 1.1
X = 1.1
X=1.1
Student B
The extension for an increase of 0.5 N is costant for student A,C and D but is not always constant for student B
1cm
6.7 cm – 1cm = 5.7 cm
2.3.
8 | P a g e
directly proportional
Newton
From the graph a load of 2.5 N causes an extension of 25mm. The total length of the spring is 50mm ( original length) + 25mm = 75mm
75mm
The force of gravity pulls it down
It has magnitude and direction
9 | P a g e
Spring 1. From the graph a force of 10N causes an extension of 12.5mm in spring 1 and an extension of 15mm in spring 2.
After point P the graph is no longer straight
P
From the graph the extensions are 19mm and 26mm. The difference in extension is 7mm
11 | P a g e
16 29 3 26 16 24 46
Error on load 3
Limit of proportionality has been reached
Find the original length of the spring. Measure the new length of the spring when it has a load. Subtract the measured length from the original length.
4.1. What is the extension of the spring? (1)
X = 20mm – 32mm = 12 mm
4.2. If the spring obeys Hooke’s Law, what does it tell you about the
extension and the load? (2)
The extension is directly proportional to the load.
4.3. What would the extension be if there was a 6N load? (2)
F = k . x
Therefore K = F/x…..
k1 = k2 therefore
F1/x1 =F2/x2
8/12 = 6 /x
8x = 48…… x = 6mm
4.4. As the weights are added the spring reaches its elastic limit. How
does this affect the spring? The extension and the load will not be directly proportional anymore
12 | P a g e